题解 | #合并两个排序的链表#

以第一个链表为基础，将第二个链表的节点插在合适的位置，但两个链表的地位不是平等的，在末尾的时候需要特殊处理。

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
#include <stdlib.h>
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode* new_head1 = (struct ListNode*)malloc(sizeof(struct ListNode*));
    new_head1->next = pHead1;
    new_head1->val = 0;
    struct ListNode* new_head2 = (struct ListNode*)malloc(sizeof(struct ListNode*));
    new_head2->next = pHead1;
    new_head2->val = 0;
    struct ListNode* cur1 = pHead1;
    struct ListNode* cur2 = pHead2;
    struct ListNode* pre1 = new_head1;
    struct ListNode* pre2 = new_head2;
    while(cur1!= NULL || cur2!= NULL)
    {
        if(cur1== NULL)
        {
            pre1->next = cur2;
            return new_head1->next;
        }
        if(cur2==NULL)
        {
            return new_head1->next;
        }
        if(cur1->val > cur2->val)
        {
            pre1->next = cur2;
            pre2->next = cur2->next;
            cur2->next = cur1;
            cur2 = pre2->next;
            pre1 = pre1->next;         
        }
        else
        {
            cur1 = cur1->next;
            pre1 = pre1->next;
        }
    }
    return new_head1->next;

}